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1、2021-2021学年春季热学习题4(答案)邢玉梅2011-05-13 1第三章 输运现象与分子动理学理论的非平衡态理论习题四：1.氧气在标准状态下的扩散系数：D =1.9?10-5m 2?s -1 求氧分子的平均自由程。解：D =13v =13?8RTM m=3D ?M m8RT=3?(1.9?10-5m 2?s -1)? 3.1416?(32?10-3kg ?m ol -1)8?(8.31J ?m ol -1?K -1)?(273K )=1.34?10-7m 2.已知氦气和氩气的原子量分别为4和40，它们在标准状态嗲的沾滞系数分别为H e =18.8?10-6N ?s ?m -2和 Ar
2、=21.0?10-6N ?s ?m -2，求：（1）氩分子与氦分子的碰撞截面之比Ar /He ；（2）氩气与氦气的导热系数之比 Ar /He ；（3）氩气与氦气的扩散系数之比D Ar /D He 。解：已知M H e =4.0g ?m ol -1,M Ar =40g ?m ol -1（1）根据 =mv 32, v =8kTm得出=mv 32=23?kTm=23?kTM m N A 由此Ar /H e =H eAr ?M Ar M H e =18.821.0?404=2.83 （2）根据=?C v ,m M m由于氮氩都是单原子分子，因而摩尔热容C v ,m 相同 Ar /He =Ar He ?
3、M He M Ar =21.0?418.8?40=0.112 （3） D =13v =PM mRTD Ar D He =Ar He ?M He M Ar =21.0?418.8?40=0.112现P 、T 都相同，3. 两个长为100cm ，半径分别为10.0cm 和10.5cm 的共轴圆筒套在一起，其间充满氢气，若氢气的沾滞系数为=8.7?10-6N ?s ?m -2 ，问外筒的转速多大时才能使不动的内筒受到107dyn 的作用力？(1dyn=10-5N)解：根据公式?/231R L f =，其中cm cm cm cm R R cm R 100L 5.00.105.10,0.10121=-=
4、-=，265107.8,10107107-?=?=m s N N dyn f 则)/8.9221s R L f （弧度=?4.两个长圆筒共轴套在一起，两筒的长度均为L ，内筒和外筒的半径分别为R 1 和 R 2，内筒和外筒分别保持在恒定的温度T 1 和T 2，且 T 1T 2，已知两筒间空气的导热系数为，试证明：每秒由内筒通过空气传到外筒的热量为：Q =2Lln R 2/R 1?(T 2-T 1)证明：由付里叶定律： dQ =-(dTdr)dSdt 则在单位时间内通过以r 为半径的柱面的热量为： Q=-(dT dr )?S =-(dTdr)2rL Q ?drr=-2LdT 由于T 1T 2 且
5、 T1、 T 2是恒量故由 T 1面传到 T 2面的热量也是一恒量，故上式两端积分得： Q 1r R 1R 2?dr =-2L dTT 1T 2?Q ln R2R 1=2L (T 1-T 2)Q =2Lln(R 2/R 1)(T 1-T 2) 5. 将一圆柱沿轴悬挂在金属丝上，在圆柱体外面套上一个共轴的圆筒，两者之间充以空气。当圆筒以一定的角速度转动时，由于空气的沾滞作用，圆柱体将受到一个力距 G . 由悬丝的扭转程度可测定此力距，从而求出空气的沾滞系数。设圆柱体的半径为R ，圆筒的半径为 R +(试证明：G=2R 3L ?/ ， 是待测的沾滞系数。证明：气体与圆柱相互作用的沾滞力为：f =?
6、(dudr)?A 只考虑平衡时的情况，可以近似地认为圆柱体不动，圆筒的转速为：v =?(R +)设紧靠圆筒的那些分子流速与圆筒转动速度相同u (R )=v因为(du dr =?(R +)=?R由牛顿沾滞定律知，圆柱体所受沾滞力邢玉梅2011-05-133f =?(du dr )?A =?R ?A 圆柱体侧面积A =2LRG T =fR =?R ?R ?2LR =2L ?R 3 6. Water flows through a fire hose of diameter 6.35cm at a rate of 0.012 m 3/s. The fire hose ends ina nozzle
7、with an inner diameter of 2.20cm. What is the speed at which the water exits the nozzle? 【answer 】the volume flow rate in unit time remains unchanged, sos m ms m v /6.31)2/022.0(/012.0223=?= 7. The effective diameter of nitrogen molecule is m 10108.3-?, to calculate the mean free path ofnitrogen mol
8、ecule at standard conditions （1atm, 0o C ）and the mean collision time for continual two times.解：=kT 2d 2P代入数据得：=(1.38?10-23J ?K -1)(273K )2?3.1416?(3.8?10-10m )2?(1.013?105Pa )=5.8?10-8m t =v=8RT M m= 5.8?10-8m1.59?(8.31J ?m ol -1?K -1)(273K )0.028Kg ?m ol -1=1.3?10-10s 8. The effective diameter of
9、oxygen molecule is 3.6?10-10m , to calculate the collision frequencyof oxygen molecules, respectively when (a) the temperature of oxygen is 300K, and the pressureis 1.0atm. (b) The temperature of oxygen is 300K, and the pressure is 6100.1-?atm.解：由 =vzz =v =8RTM m kT 2d 2P代入数据得： z 1=6.3?109s -1 z 2=P
10、 2P 1z 1=10-6?6.3?109s -1=6.3?103s -19. The mean free path of some kind of gas molecule is 2.63?10-7m when the temperature is25, (a) to calculate gas pressure when the effective diameter of gas molecule is 2.6?10-10m . (b) To calculate the collision times with other gas molecules in 1m distance.解：（1
11、）由=kT2d 2P得： P=kT2d 2代入数据得：P=(1.38?1023J?K-1)(298.15K)2?3.1416?(2.6?10-10m)2?(2.63?10-7m)=5.21?104J?m-3=0.521atm（2）分子走路程碰撞次数N=L=1.0m2.63?10-7m=3.8?106（次）10.Oil at 20o C (=888kg/m3and =0.8kg/m?s)is flowing steadily through a 30m longcylinder pipe with a diameter of a 5cm. During the flow, the pressur
12、e at the pipe inlet and exit are measured to be 823 KPa nd 96 KPa, respectively. Determine the flow rate of oil through the pipe assuming the pipe is (1) horizontal, (2) inclined 15o upward, (3) inclined 15o downward.Also verify that the flow through the pipe is laminar.解：错误！未找到引用源。The flow rate of all three cases can be determined from(1)=0,So sin=0(2)=15,So sin=0.2588(3)=-15,So sin=-0.2588
